# Difference between revisions of "Star formation"

### Star Formation

\Stars form out of cold gas in galaxies. The gas in a galaxy like the Milky Way is in a flat sheet, not unlike the dimensions of a stack of a few pieces of paper. For the Milky Way, radial size of the disk is roughly 10 kiloparsecs, where a parsec is roughly three lightyears. The height of the disk is some hundreds of parsecs, so roughly one tenth, or less, than the radial extent. The radial extent is a trillion times larger than the Sun, which gives a hint of the problem of star formation. How does gas get taken from large scales and turned in to a (relatively) small object like a star? Another way to see this issue is to look at the mean density of the galaxy, and the mean density of the Sun. The interstellar medium (ISM) has a mean density of roughly 1 particle per cubic centimeter. The Sun has a density of one gram per cubic centimeter. Since most things are hydrogen, that means the Sun is ${\displaystyle 10^{24}}$ times denser than the ISM. \\The ISM actually comes in a wide range of temperatures and densities, so this single number does not tell the whole picture. All of these phases are roughly in pressure equilibrium though, as there would be forces between the phases acting to bring them to equilibrium were they not actually in equilibrium. The phases are

 Name Number density (n) Temperature (T) Constituent Hot Ionized Medium ${\displaystyle 10^{-2}}$ cm${\displaystyle ^{-3}}$ ${\displaystyle 10^{6}}$ K Ionized hydrogen Warm Ionized Medium ${\displaystyle 1}$ cm${\displaystyle ^{-3}}$ ${\displaystyle 10^{4}}$ K Ionized hydrogen Cold Neutral Medium ${\displaystyle 10^{2}}$ cm${\displaystyle ^{-3}}$ ${\displaystyle 100}$ K Neutral atomic hydrogen Molecular Clouds ${\displaystyle >10^{2}}$ cm${\displaystyle ^{-3}}$ ${\displaystyle 10}$ K Molecular hydrogen

The top two columns have most of the volume, and the bottom two most of the mass. With number densities in excess of average, the bottom two are also rare. They are where stars actually form in a galaxy, since they are slightly more dense than the rest of the ISM. Even so, they are much less dense than a star. How do we get a star then? Gravitational collapse. \\Imagine a cloud with mass ${\displaystyle M}$, radius ${\displaystyle R}$, density ${\displaystyle \rho }$, and temperature ${\displaystyle T}$. The gravitational potential energy of the cloud is

${\displaystyle U\sim -{\frac {GM^{2}}{R}}.\,\!}$

Compare this to the thermal energy of the cloud, which is

${\displaystyle K\sim NkT\sim {\frac {M}{m_{p}}}kT.\,\!}$

The condition for collapse then is that the potential energy be greater than the kinetic energy,

${\displaystyle |U|>K.\,\!}$

Plugging in our values for the cloud,

${\displaystyle {\frac {GM^{2}}{R}}>{\frac {M}{m_{p}}}kT,\,\!}$
${\displaystyle M>{\frac {RkT}{Gm_{p}}},\,\!}$
${\displaystyle \rho \sim {\frac {M}{R^{3}}},\,\!}$
${\displaystyle M>{\frac {M^{1/3}}{\rho ^{1/3}}}{\frac {kT}{m_{p}G}},\,\!}$
${\displaystyle M>\left({\frac {k}{m_{p}G}}\right)^{3/2}{\frac {T^{3/2}}{\rho ^{1/2}}}.\,\!}$

If we put in some numbers, we find that

${\displaystyle M_{J}\approx 50M_{\odot }\left({\frac {T}{10K}}\right)^{3/2}\left({\frac {n}{100{\rm {\;cm}}^{-3}}}\right)^{-1/2}.\,\!}$

Only when ${\displaystyle M>M_{J}}$ will collapse occur. This indicates that only the cool, dense gas can form stars. For the hotter components, the Jeans mass is of order the mass of all the gas in the galaxy, so it is not in danger of collapse. We can also define a Jeans length, by rearrangement of the above. We find

${\displaystyle R_{J}=\left({\frac {k}{Gm_{p}}}\right)^{1/2}{\frac {T^{1/2}}{\rho ^{1/2}}}.\,\!}$

Again putting in numbers, we find

${\displaystyle R_{J}\approx 3{\rm {\;pc}}\left({\frac {T}{10K}}\right)^{1/2}\left({\frac {n}{100{\rm {\;cm}}^{-3}}}\right)^{-1/2}.\,\!}$

This is much bigger than the size of a star (about one hundred million times larger). So there is a lot between collapse and the formation of a star. How long does it take for collapse to occur? Well, the free fall time is

${\displaystyle t\sim {\sqrt {\frac {1}{G\rho }}}.\,\!}$

This is roughly ten million years for the density of a molecular cloud. \\What happens once collapse begins? Do we get lots of small stars, or one very massive stars? Force balance and energy losses determine the fate of a cloud. We can write down the acceleration as

${\displaystyle \rho a=-{\frac {dP}{dr}}-\rho {\frac {GM}{r^{2}}}.\,\!}$

What happens then depends on whether pressure or gravity wins during collapse. Order of magnitude, the pressure term scales as

${\displaystyle {\frac {P}{R}}\sim {\frac {nkT}{R}}\propto {\frac {TM}{R^{4}}}.\,\!}$

The gravity term scales as

${\displaystyle \rho {\frac {GM}{r^{2}}}\propto {\frac {M^{2}}{R^{5}}}.\,\!}$

In order to compare these, we need to know how the temperature changes as the cloud contracts. We can look at two cases here. One is when the mean free path of photons is larger than the cloud, which means they are able to effectively carry energy out of the cloud. That means the temperature is roughly constant, meaning that pressure will scale as ${\displaystyle R^{-4}}$ while gravity will scale as ${\displaystyle R^{-5}}$. Thus gravity will get stronger relative to pressure, and the collapse of the cloud will run away initially. What is happening to the Jeans’ mass as this occurs? Well, the density is increasing, the temperature is constant, and the Jeans’ mass decreases. Thus it will be possible for the cloud to break up in to smaller chunks as gravitational collapse proceeds, because individual clumps within the cloud can break apart in to separate clumps that are collapsing on their own. This is called gravitational fragmentation. \\The second case in collapse is when the cloud becomes optically thick, or the mean free path is small relative to the size of the clump. When photons cannot escape, energy cannot easily leave the cloud, and the collapse is adiabatic. How does the temperature change in an adiabatic process? From earlier, ${\displaystyle T\propto \rho ^{\gamma -1}}$, or ${\displaystyle T\propto \rho ^{2/3}\propto M^{2/3}R^{2}}$. The Jeans’ mass has dependence ${\displaystyle M_{J}\propto T^{3/2}\rho ^{-1/2}\propto \rho ^{1/2}}$. Thus the Jeans’ mass increases, and further fragmentation is halted. In terms of the pressure, we have the pressure force increasing as ${\displaystyle R^{-6}}$, which is a larger dependence than the gravity. Thus the pressure force increases more rapidly than gravity does, and the cloud will come in to hydrostatic equilibrium. At this point, the clump is much larger than a star, but it is getting close now that it is in hydrostatic equilibrium. Observationally, a typical star is about half a solar mass. For stars more massive than about a third of a solar mass, the distribution of solar masses (known as the Salpeter initial mass function) is

${\displaystyle N(M)\propto M^{-1.35}.\,\!}$

Our cloud in HE with no fusion will contract as it loses energy from radiating away energy at the point where it becomes optically thin. The cloud will then undergo Kelvin-Helmholtz contraction. Since it is in HE, the Virial Theorem holds. Then

${\displaystyle E_{tot}={\frac {U}{2}}.\,\!}$

Then the luminosity is

${\displaystyle L=-{\frac {dE_{tot}}{dt}}=-{\frac {1}{2}}{\frac {dU}{dt}}.\,\!}$

This object is fully convective (which we won’t fully justify, but is not surprising given how cool the cloud is). Using an ${\displaystyle n=3/2}$ polytrope, we know then

${\displaystyle U=-{\frac {6}{7}}{\frac {GM^{2}}{R}}\,\!}$

Then the luminosity, in terms of the mass and radius, is

${\displaystyle L={\frac {3}{7}}{\frac {GM^{2}}{R^{2}}}\left|{\frac {dR}{dt}}\right|.\,\!}$

We also know that for fully convective objects,

${\displaystyle L=0.2L_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{4/7}\left({\frac {R}{R_{\odot }}}\right)^{2}.\,\!}$

We can solve for how the contraction proceeds by using this known convective luminosity. If we set the two expressions equal to one another, we find

${\displaystyle {\frac {3}{7}}{\frac {GM^{2}}{R^{2}}}\left|{\frac {dR}{dt}}\right|=0.2L_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{4/7}\left({\frac {R}{R_{\odot }}}\right)^{2}.\,\!}$

If we solve for the radius as a function of time, we find

${\displaystyle R\sim R_{\odot }\left({\frac {t}{2\times 10^{7}{\rm {\;years}}}}\right)^{-1/3}\left({\frac {M}{M_{\odot }}}\right)^{1/2}.\,\!}$

We can also solve for the luminosity, finding

${\displaystyle L\sim L_{\odot }\left({\frac {t}{2\times 10^{7}{\rm {\;years}}}}\right)^{-2/3}\left({\frac {M}{M_{\odot }}}\right)^{11/7}.\,\!}$

Initially, the radius and luminosity of the star are both much larger than that of the Sun. As the star (or, pre-star) contracts, it will move down the Hayashi line at roughly constant temperature. At some point though, photons will take over the energy transport, at least for stars more massive than one third of a solar mass. We can find this point by equating the expressions for luminosity carried out by photons and luminosity carried out by convection. When a star crosses that point, it will cease moving down the Hayashi line, and instead move across the HR diagram until it reaches its main sequence position. We can solve for this time, and find that the time at which photons dominates is

${\displaystyle t_{photons}\sim 10^{6}{\rm {\;years}}\left({\frac {M}{M_{\odot }}}\right)^{-15/7}.\,\!}$

Thus for massive stars, photons carry out the energy for the majority of their pre main sequence lifetime. \\Finally, when does contraction stop? When fusion starts. Or, more precisely, when fusion can generate the energy that leaves the star, so that it is not necessary to contract any further. This takes roughly twenty million years for a star of the mass of the Sun. This is, not coincidentally, of order the Kelvin-Helmholtz time.